Discrete Probability Distribution
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Last updated on November 26, 2025

Probability is the possibility that an event will occur. A discrete probability distribution describes the likelihood of outcomes for individually countable variables. In this topic, we will learn more about discrete probability distribution, its types, application, and many more.

What is a Discrete Probability Distribution?

The discrete probability distribution is a simple way to show the different chances of different outcomes when you can count them one by one. It gives the different values of a random variable along with its different probabilities. A discrete probability distribution contrasts with a continuous distribution, where outcomes can take any value along a continuum, as the outcome falls anywhere on a continuum. The types of discrete probability are Binomial, Poisson, and Bernoulli distributions. If a probability distribution is said to be a discrete probability distribution, it should follow these two conditions : 
 

  • 0 ≤ P(X = x) ≤ 1, which means the discrete random variable, X should be on the exact value, x, and it should be between 0 and 1
  • ΣP (X = x) = 1, that is, the sum of all probabilities should be 1.
     

Example:

Imagine you flip a fair coin 3 times, and you want to count the Number of Heads (X).

This is a perfect example because:

  • It is discrete: You can only get 0, 1, 2, or 3 heads. You can't get 1.5 heads.
  • It is finite: There is a limited number of possible outcomes.

Step 1: List all possible outcomes

There are 8 total ways the coins can land (\(2^3 = 8\)):

  • 3 Heads: HHH
  • 2 Heads: HHT, HTH, THH
  • 1 Head: HTT, THT, TTH
  • 0 Heads: TTT.

Step 2: Create the Distribution Table

Now, we count the probability for each number of heads.

Number of Heads (x) Frequency (How many ways?) Probability P(X=x) 0 1 (TTT only) 1/8 (or 0.125) 1 3 (HTT, THT, TTH) 3/8 (or 0.375) 2 3 (HHT, HTH, THH) 3/8 (or 0.375) 3 1 (HHH only) 1/8 (or 0.125)

Discrete Probability Distribution Formula

Here are the formulas for the PMF and CDF of Discrete Probability Distribution.


PMF of Discrete Probability Distribution


The PMF (Probability Mass Function) gives the probability that a discrete random variable X is exactly equal to a specific value x.


Formula:
\(f(x) = P(X = x)\)


Example:
Let X be the number of Girls in a family with 3 children.


Possible Outcomes (\(2^3 = 8\) total outcomes):

  • GGG \(\rightarrow\) 3 Girls
  • GGB, GBG, BGG \(\rightarrow\) 2 Girls
  • GBB, BGB, BBG \(\rightarrow\) 1 Girl
  • BBB \(\rightarrow\) 0 Girls
     

The PMFs are given as:

  • \(P(X=0) = \frac{1}{8} \rightarrow\) (Only BBB)
  • \(P(X=1) = \frac{3}{8} \rightarrow\) (GBB, BGB, BBG)
  • \(P(X=2) = \frac{3}{8} \rightarrow\) (GGB, GBG, BGG)
  • \(P(X=3) = \frac{1}{8} \rightarrow\) (Only GGG)
     

Total Probability:
\(\frac{1}{8} + \frac{3}{8} + \frac{3}{8} + \frac{1}{8} = 1\)


Properties of PMF:

  1. The sum of all probabilities must equal 1: \(\sum P(X=x) = 1\)
  2. Probabilities cannot be negative: \(P(X=x) \ge 0\)

CDF of Discrete Probability Distribution


The CDF (Cumulative Distribution Function) gives the probability that the random variable X is less than or equal to a specific value x.


Formula:
\(F(x) = P(X \le x)\)


Example:
Let a random variable X be the score obtained on a biased (weighted) 4-sided die. We want to find the probability \(P(1 < X \le 3).\)


Data Table:

X (Score)

PMF P(X=x)

CDF F(x)=P(X≤x)

1 0.1 0.1 2 0.4 0.1 + 0.4 = 0.5 3 0.3 0.5 + 0.3 = 0.8 4 0.2 0.8 + 0.2 = 1.0

Calculation for \(P(1 < X \le 3)\):


Using the CDF property:
\(P(a < X \le b) = F(b) - F(a)\)


Therefore:
\(P(1 < X \le 3) = F(3) - F(1)\)

\(P(1 < X \le 3) = 0.8 - 0.1 = 0.7\)


(Verification using PMF): \(P(1 < X \le 3)\) means we want outcomes 2 and 3.
\(P(X=2) + P(X=3) = 0.4 + 0.3 = 0.7\)

What are the Types of Discrete Probability Distributions

Now let’s discuss the types of discrete probability. The types are;

  • Binomial Distribution
    Binomial probability is where there is a probability of getting two outcomes. Here after repetitive trails the data is collected in one of the two forms, into either success or failure. For instance, when flipping a coin, the probability of getting heads. 
     
  • Bernoulli Distribution
    It is similar to binomial probability, as it also has two possible outcomes, with only one trial. Even here, the possible outcomes will be success or failure. 
     
  • Multinomial Distribution
    In multinomial distribution, the probability of getting more than two outcomes with multiple counts. For example, a bowl filled with three different colored coins like red, blue, and green.
     
  • Poisson Distribution
    It is a type of probability that tells how a certain number of events happen in a fixed amount of time. For example, a Poisson distribution can be used for counting how many times a bus arrives at a stop in a fixed amount of time.
     
  • Geometric Distribution
    It is a type of probability that tells you how many independent trials are needed to achieve the very first success. For example, a Geometric distribution can be used for counting how many job applications you need to submit until you finally get your first offer.
     

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Mean of Discrete Probability Distribution

The Mean (also called the Expected Value) of a discrete probability distribution is the weighted average of all possible values that the random variable can take. It represents the long-term average result if you were to repeat the experiment many times.
 

Formula:
The mean \mu (or E(X)) is calculated by multiplying each possible outcome x by its probability P(X=x) and then adding them all together:


\(\mu = E(X) = \sum [x \cdot P(X=x)]\)

Example:
Let a random variable X be the cash prize won in a charity raffle game.


Possible Outcomes:

  • $0 (No win)
  • $10 (Small prize)
  • $50 (Grand prize)

Calculation Table: We multiply each outcome (x) by its probability (P(x)) to find the “weighted” value.

Outcome (x)

Probability P(X=x)

Weighted Value [x⋅P(x)]

0 0.80 \(0 \cdot 0.80 = \mathbf{0}\) 10 0.15 \(10 \cdot 0.15 = \mathbf{1.5}\) 50 0.05 \(50 \cdot 0.05 = \mathbf{2.5}\)

Total Mean:
\(E(X) = 0 + 1.5 + 2.5 = 4.0\)

Interpretation:
The expected value is $4. This means that if you played this game thousands of times, you would win an average of $4 per game, even though you can never actually win exactly $4 in a single turn.

Variance of Discrete Probability Distribution

The Variance measures how “spread out” or dispersed the numbers are from the mean (expected value). A high variance means the values are widely scattered; a low variance means they are clustered closely around the average.


Formula:
There are two common ways to calculate it.

  1. The Definition Formula: The weighted average of the squared differences from the mean.

    \(\sigma^2 = Var(X) = \sum [(x - \mu)^2 \cdot P(x)]\)

  2. The Computational Formula (Often Easier): The average of the squares minus the square of the average.

    \(\sigma^2 = E(X^2) - [E(X)]^2 = \sum [x^2 \cdot P(x)] - \mu^2\)


Example:
Let a random variable X be the number of daily complaints received by a service center.


Possible Outcomes:

  • 0 complaints (Quiet day)
  • 1 complaint (Normal day)
  • 2 complaints (Busy day)

Probabilities:

  • P(X=0) = 0.1
  • P(X=1) = 0.6
  • P(X=2) = 0.3

Step 1: Find the Mean (\mu)
\(\mu = (0 \cdot 0.1) + (1 \cdot 0.6) + (2 \cdot 0.3)\)
\(\mu = 0 + 0.6 + 0.6 = \mathbf{1.2}\)
\(\mu^2 = 1.2^2 = 1.44\)


Step 2: Calculate \(E(X^2)\)
We need to square each outcome first, then multiply by its probability

Outcome (x)

Probability P(x)

Squared Outcome \((x^2)\) Weighted Square\(E(X^2) = [x^2⋅P(x)]\) 0 0.1 \(0^2 = 0\) \(0 \cdot 0.1 = \mathbf{0}\) 1 0.6 \(1^2 = 1\) \(1 \cdot 0.6 = \mathbf{0.6}\) 2 0.3 \(2^2 = 4\) \(4 \cdot 0.3 = \mathbf{1.2}\) \(\sum\)     \(0+0.6+1.2=1.8\)

Step 3: Apply the Formula
Now we subtract the square of the mean from the sum we just found.

Variance(\(\sigma^2\)) 
\(\sigma^2 = E(X^2) - \mu^2\)
\(\sigma^2 = 1.8 - 1.44\)
\(\sigma^2 = \mathbf{0.36}\)

Standard deviation(\(\sigma\))
\(\sigma = \sqrt{0.36} = \mathbf{0.6}\)

How to Calculate Discrete Probability Distribution

Here, we will discuss how to calculate a discrete probability distribution. To find a discrete probability distribution, the probability of mass function is required. Follow these steps to find the probability;

Step 1: Identify the sample, set of all possible outcomes of an experiment. 

Step 2: Then you have to find the discrete random variable (x). It is a function assigning a numerical value to each outcome.


Step 3: Identifying the possible value of X. 


Step 4: Finding the probability of each outcome by using the formula; 

P (X = x) = Number of favorable outcomes/Total number of possible outcomes. 

Step 5: Use a table to organize the results

Discrete Distribution vs. Continuous Distribution

Now let’s learn the difference between the discrete and continuous distribution. 

Discrete Distribution

Continuous Distribution

The probability distribution for countable values

The probability distribution for measurable values

Here, the type of variable is discrete 

Here, the type of variable is continuous 

The graph of discrete distribution is a bar

The graph of continuous distribution is a curve 

Tips and Tricks to Master Discrete Probability Distribution


Discrete probability distribution is a complex topic to get a grasp on. In this section, we will discuss some tips and tricks to master Discrete probability distribution.
 

  • Organize your data in a simple table with two columns: One for the random variable (X) and one for its probability P(X).
     
  • Practice with real-life scenarios: Apply discrete probability to everyday examples like tossing coins, quality checks, or customer arrivals.
     
  • Identify the type of distribution: Understand whether the problem is about a fixed number of trials, rare events, or waiting for the first success. Correctly identifying the type helps you approach it the right way.
     
  • Understand conditions: Check the assumptions behind each distribution, like independence of events or whether outcomes are countable. This prevents applying the wrong distribution.
     
  • Use expected patterns: Think about what results are reasonable based on the situation. This helps spot errors in calculations or assumptions.

Common mistakes and How to Avoid Them in Discrete Probability Distribution

Now let’s learn a few common mistakes and ways to avoid them to master discrete probability distribution. 
 

Real-Life Applications of Discrete Probability Distribution

In real-life discrete probability distribution are used to find the probability for countable values, now lets few real-world applications of it;

  • For quality control in manufacturing, we use binomial distribution to find the number of defective products in a batch based on the random checks. 
  • When playing the games of chances, like for example cards, snake and ladder, etc. Probability helps you figure out how likely you are to win. 
  • The number of calls received in a call center is predicted using the Poisson distribution.
  • To identify whether an email is a spam or not using the Bernoulli distribution.
     
  • Insurance companies use the normal distribution to predict risk and estimate claims.
     

Problem 1

A factory produces light bulbs, and 5% of them are defective. If a random sample of 8 bulbs is taken, what is the probability that exactly 2 bulbs are defective?

Okay, lets begin

The probability of getting 2 bulbs are 0.0515 or 5.15%.
 

Explanation

Here, to find the probability we use the equation, P(X = k) = \(\binom{n}{k}\) pk(1 - p)n-k


According to the problem, 


n = 8


k = 2


p = 0.05


Substituting the values, 


P(X = 2) = \(\binom{8}{2}\) (0.05)2(1 - 0.05)8-2


P(X = 2) = \(\binom{8}{2}\) (0.05)2( 0.95)6

The value of C \(\binom{8}{2}\)= 28, (0.05)2 = 0.0025, and (0.95)6 = 0.7358


So, the value of P(X = 2) = 28 × 0.0025 × 0.7358 = 0.0515.

Well explained 👍

Problem 2

A basketball player has a 40% probability of making a free throw. What is the probability that their first successful shot happens on the third attempt?

Okay, lets begin

The probability here is 0.144 or 14.4%.
 

Explanation

Here the probability is of geometric distribution


So, P(X = k) = (1 - p)(k-1) × p.


Here, p = 0.4 and k = 3


Therefore, P(X = 3) = (0.6)2 × 0.4 


= 0.36 × 0.4 = 0.144.

Well explained 👍

Problem 3

A deck has 52 cards, including 4 Aces. If 5 cards are drawn randomly, what is the probability of getting exactly 2 Aces?

Okay, lets begin

Probability ≈ 0.0399 or 3.99%.

Explanation

Here, the probability is calculated using  \(P (X = k) = \dfrac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}.\)


Here, 


N = 52


K = 4


k = 2


n = 5

So P(X = 2) =\( \dfrac{\binom{4}{2} \binom{52 - 4}{5 - 2}}{\binom{52}{5}}.\)


P(X = 2) = 0.0399 or 3.99%.
 

Well explained 👍

Problem 4

A multiple-choice quiz has 10 questions, each with 4 answer choices. A student guesses randomly on each question. What is the probability of getting exactly 3 correct answers?

Okay, lets begin

The probability is 0.250 or 25%.

Explanation

Here, to find the probability we use the equation, P(X = k) = C \(\binom{n}{k}\) pk (1 - p)(n - k)


According to the problem, 


n = 10


k = 3


p = ¼ = 0.25


Substituting the values, 


P(X = 3) = \(\binom{10}{3}\)(0.25)2(1 - 0.25)10 - 3


P(X = 3) = \(\binom{10}{3}\) (0.25)2(0.75)7

Here, C\(\binom{10}{3}\) = 120


0.252 = 0.015625


0.757 = 0.1335

P(X = 3) ≈ 120 × 0.015625 × 0.1335 ≈ 0.250.

Well explained 👍

Problem 5

A factory produces screws with a 2% defect rate. What is the probability that the first defective screw appears on the 5th inspection?

Okay, lets begin

The probability is 0.01845 or 1.845%.
 

Explanation

P(X = 5) = (1 - p)(k-1) × p.


Here, p = 0.02


K = 5


P(X=5) = (1 − p)(k-1)p = (0.98)4 × 0.02


As 0.984 = 0.9224


 P(X = 5) = 0.9224 × 0.02 = 0.01845.

Well explained 👍

FAQs on Discrete Probability Distribution

1.What is a discrete probability distribution?

A discrete probability distribution is one type of probability that shows different chances of different countable outcomes of an event that are to happen.
 

2.What is the difference between discrete probability distribution and continuous probability?

Discrete probability distribution deals with counted numbers whereas continuous probability deals with the measurements
 

3.What are the real-life applications of discrete probability distribution?

Discrete probability distribution is used in real-life situations to find customers' arrivals, phone calls, chances of winning a game and so on. 
 

4.What are the types of discrete probability distribution?

Bernoulli distribution, binomial distribution, Poisson distribution, and multinomial distribution are the four types of probability distributions. 
 

5.What are the conditions that a discrete probability distribution follows?

Discrete probability follows two conditions, that is p(x) ≥ 0 and Σp(x) = 1
 

Jaipreet Kour Wazir

About the Author

Jaipreet Kour Wazir is a data wizard with over 5 years of expertise in simplifying complex data concepts. From crunching numbers to crafting insightful visualizations, she turns raw data into compelling stories. Her journey from analytics to education ref

Fun Fact

: She compares datasets to puzzle games—the more you play with them, the clearer the picture becomes!